I kwesibalo indiza: indlela yokwenza? Izinhlobo indiza zibalo

Isikhala indiza kungachazwa ngezindlela ezahlukene (owodwa ichashaza nefremu ye-vector, vector kanye amaphuzu amabili, amaphuzu amathathu, njll). Kungenxa nalesi engqondweni, ezothando indiza ungaba izinhlobo ezahlukene. Futhi ngaphansi kwezimo ezithile indiza kungenzeka okufanayo, perpendicular, intersecting, njll Ngalolu futhi ngeke ukhulume kule athikili. Sizofunda ukwenza kwesibalo jikelele endizeni futhi hhayi kuphela.

Ifomu evamile kwesibalo

Ake sithi R isikhala _3, okuyinto has a elingunxande Ukudidiyela uhlelo XYZ. Sichaza α vector, okuyinto bayokhululwa esizoqala O. Through ekupheleni α Vector ukudweba indiza P okuyinto perpendicular ke.

Lumelela P at ongumashiqela iphuzu Q = (x, y, z). Irediyasi Vector iphuzu Q incwadi uphawu p. Ubude vector kulingana p α = IαI futhi Ʋ = (cosα, cosβ, cosγ).

Le yunithi vector, okuyinto iqondiswa ohlangothini njengoba Vector α. α, β futhi γ - kukhona engeli ukuthi akhiwa phakathi Vector futhi iziqondiso ezinhle Ʋ isikhala izimbazo x, y, z ngokulandelana. I Iwumbono iphuzu ku vector QεP Ʋ kuyinto njalo okuyinto ilingana p (p, Ʋ) = p (r≥0).

I kwesibalo ngenhla okunengqondo uma p = 0. Okuwukuphela n indiza kule ndaba, kungaba ukuwela iphuzu O (α = 0), okuyinto umsuka, futhi iyunithi Vector Ʋ, ukhululeka iphuzu O kuyoba perpendicular P, nakuba isiqondiso sayo, okusho ukuthi Ʋ Vector kunqunywa kuze kube uphawu. equation Langaphambilini kuyinto indiza yethu P oluboniswa ifomu le-vector. Kodwa uma sicabangela ukuthi izixhumanisi salo:

P mkhulu noma elilingana 0. Simfumene kwesibalo indiza e ifomu evamile.

I kwesibalo jikelele

Uma kwesibalo e izixhumanisi wande noma iyiphi inombolo ukuthi akuyona ilingane no-zero, sithola okulingana kwesibalo kule ukuthi ichaza kakhulu indiza. Lizoba ifomu elilandelayo:

Lapha, A, B, C - inombolo kanyekanye sihluke zero. Lokhu equation ubizwa ngokuthi kwesibalo yefomu jikelele indiza.

I zibalo izindiza. Ezimweni Special

I kwesibalo ungahlukaniswa ngokuvamile ube Ushintshe nezimo ezengeziwe. Ake ucabangele ezinye zazo.

Ucabange ukuthi Coefficient A 0. Lokhu kubonisa ukuthi parallel indiza eksisi Ox kusengaphambili. Kulokhu, ngesimo ezothando sishintsha: Wu + CZ + D = 0.

Ngokufanayo, indlela equation kanye kuzohluka nge kwale mibandela elandelayo:

• Okokuqala, uma B = 0, izinguquko equation ukuze Axe + CZ + D = 0, okwakuyoholela ukhombise impindamqondo ukuze eksisi Oy.
• Okwesibili, uma C = 0, ezothando uguqulwe Axe + ngu + D = 0, okungukuthi cishe parallel eksisi predetermined Oz.
• Okwesithathu, uma D = 0, ezothando lizovela Axe + ngu + CZ = 0, okuyinto kungasho ukuthi indiza ephambana O (umsuka).
• Okwesine, uma A = B = 0, izinguquko equation ukuze CZ + D = 0, okuyinto iyohlala kuze impindamqondo Oxy.
• Okwesihlanu, uma B = C = 0, ezothando iba Axe + D = 0, okusho ukuthi indiza parallel Oyz.
• Sixthly, uma A = C = 0, ezothando kuthatha ifomu Wu + D = 0, ngamanye amazwi, azobika kuya Oxz impindamqondo.

Ifomu le-equation ngezigaba

Esimweni lapho izinombolo A, B, C, D sihluke zero, ngesimo kwesibalo (0) kungenzeka kanje:

x / a + y / b + z / c = 1,

lapho a = -D / A, b = -D / B, c = -D / C.

Sithola ngenxa kwesibalo indiza yaba yizicucu. Kumele kuqashelwe ukuthi le indiza ngeke aphambana x-axis endaweni nge izixhumanisi (a, 0,0), Oy - (0, b, 0), kanye Oz - (0,0, s).

Banikezwe kwesibalo x / a + y / b + z / c = 1, akunzima ukubona ngehlo lengqondo i isihlobo indiza ukubekwa kusistimu predetermined Ukudidiyela.

Izixhumanisi Vector evamile

I evamile Vector n kuya P indiza has izixhumanisi ukuthi kukhona okuza of the equation jikelele we indiza, isb n (A, B, C).

Ukuze ukwazi ukubona izixhumanisi n evamile, kwanele ukuba wazi kwesibalo jikelele unikezwa indiza.

Uma usebenzisa kwesibalo ngezigaba, okuyinto ine efomu x / a + y / b + z / c = 1, njengoba uma usebenzisa kwesibalo jikelele kungenziwa izixhumanisi iyiphi Vector evamile ebhaliwe indiza enikeziwe: (1 / a +1 / b +1 / c).

Kufanele kuqashelwe ukuthi Vector evamile isize ekuxazululeni izinkinga ezihlukahlukene. Izinkinga evamile owakhiwa izindiza ubufakazi perpendicular noma parallel, umsebenzi ekutholeni engele phakathi izindiza noma engele phakathi izindiza kanye imigqa eqondile.

Thayipha ngokuvumelana kwesibalo indiza kanye izixhumanisi iphuzu Vector evamile

A nonzero Vector n, perpendicular indiza unikezwa, okuthiwa ezivamile (evamile) ukuze indiza kusengaphambili.

Ake sithi Ukudidiyela isikhala (a elingunxande Ukudidiyela uhlelo) Oxyz setha:

• Mₒ iphuzu nge izixhumanisi (hₒ, uₒ, zₒ);
• zero Vector n = A * i + B * j + C * k.

Udinga ukwenza kwesibalo indiza ukuthi sidlula Mₒ iphuzu perpendicular n evamile.

Esikhaleni sikhetha iliphi iphuzu ngokungenasizathu futhi selisho M (x, y, z). Ake vector engaba iphuzu M ngamunye (x, y, z) kuyoba r = x * i + y * j + z * k, futhi engaba Vector ye Mₒ iphuzu (hₒ, uₒ, zₒ) - rₒ = hₒ * i + uₒ * j + zₒ * k. Iphuzu M utawuba indiza unikezwa, uma MₒM Vector perpendicular Vector n. Sibhala isimo orthogonality ukusebenzisa umkhiqizo scalar:

[MₒM, n] = 0.

Njengoba MₒM = r-rₒ, ezothando Vector indiza izobukeka kanje:

[R - rₒ, n] = 0.

Lokhu kwesibalo Ungase futhi ube omunye ukuma. Ngenxa yale njongo, property umkhiqizo scalar, futhi iguqulwe ohlangothini lwesobunxele ezothando. [R - rₒ, n] = [r, n] - [rₒ, n]. Uma [rₒ, n] ezikhonjiswe yemagama, sithola ezothando ezilandelayo: [r, n] - a = 0 noma [r, n] = s, azwakalisa lo ukuqinisela we emisebenzi ku vector ejwayelekile engaba-zithwala amaphuzu unikezwa ukuthi eyakhe indiza.

Manje ungathola Ukudidiyela uhlobo ukuqoshwa indiza Vector yethu kwesibalo [r - rₒ, n] = 0. Kusukela r-rₒ = (x-hₒ) * i + (y uₒ) * j + (z-zₒ) * k, futhi n = A * i + B * j + C * k, sinazo:

It kuvela ukuthi sinalo kwakheka kwesibalo is indiza edabula iphuzu perpendicular n evamile:

A * (x hₒ) + B * (y uₒ) S * (z-zₒ) = 0.

Thayipha ngokuvumelana kwesibalo indiza kanye izixhumanisi amaphuzu amabili we collinear indiza Vector

Sichaza amaphuzu amabili ngokungenasizathu M '(x, y, z) kanye M "(x", y ", z"), kanye vector (a', a ', a ‴).

Manje singakwazi ukubhala kwesibalo predetermined indiza lapho sidlula iphuzu M akhona futhi M ", futhi iphuzu ngalinye izixhumanisi M (x, y, z) parallel a vector inikezwe.

Ngakho M'M zithwala x = {x ', y-y'; ZZ '} futhi M "M = {x" -X', y 'y'; z "-z '} kufanele kube coplanar ne Vector a = (a ', a', a ‴), okusho ukuthi (M'M M "M, a) = 0.

Ngakho kwesibalo kwethu indiza emkhathini izobukeka kanje:

Uhlobo indiza equation, ukweqa amaphuzu amathathu

Ake sithi sinalo amaphuzu amathathu: (x, y, z), (x, y, z), (x ‴ Have ‴, z ‴), okuyinto ezakho line efanayo. Kuyadingeka ukuba abhale kwesibalo indiza edabula amaphuzu amathathu ecacisiwe. geometry ithiyori uthi lolu hlobo indiza likhona, kungcono nje yedwa. Njengoba lokhu indiza ephambana iphuzu (x, y, z), injalo kwesibalo kungaba:

Lapha, A, B, no-C bengafani zero ngesikhathi esifanayo. Futhi indiza unikezwa ephambana amaphuzu amabili ngaphezulu (x ", y", z ") kanye (x ‴, y ‴, z ‴). Kule ndaba kufanele kwenziwe lolu hlobo izimo:

Manje singakwazi ukwakha uhlelo iyunifomu of zibalo (lwento) nge angaziwa u, v, w:

Esimweni sethu x, y noma z imele iphuzu ngokungenasizathu lapho kwanelisa kwesibalo (1). Uma ucabangela kwesibalo (1) kanye nohlelo zibalo (2) no- (3) uhlelo of zibalo kukhonjisiwe sibalo ngenhla, vector wanelisa N (A, B, C) okuyinto nontrivial. Kungenxa yokuthi determinant kohlelo zero.

Isibalo (1) ukuthi sinezinto, lena kwesibalo indiza. 3 iphuzu eyaphi ngempela, futhi kulula ukuhlola. Ukuze wenze lokhu, thina ukwandisa determinant yi izakhi emgqeni wokuqala. Izakhiwo ezikhona determinant kulandela ukuthi indiza yethu kanyekanye ephambana iphuzu emithathu ekuqaleni predetermined (x, y, z), (x ", y", z "), (x ‴, y ‴, z ‴). Ngakho sanquma komsebenzi phambi kwethu.

engela Dihedral phakathi izindiza

engela Dihedral kuyinto ukuma ngekwendzawo weJiyomethri ezakhiwe ezimbili nengxenye-ezindizeni ezibangelwa umugqa oqondile. Ngamanye amazwi, ingxenye isikhala lapho kunomkhawulo isigamu-ezindizeni.

Ake sithi ube ezimbili indiza ne zibalo ezilandelayo:

Siyazi ukuthi N Vector = (A, B, C) futhi N¹ = (¹, H¹, S¹) ngokuvumelana izindiza predetermined kukhona perpendicular. Kule ndaba, i-engeli φ phakathi zithwala N futhi N¹ engela alinganayo (dihedral), esemgwaqweni phakathi kwalezi izindiza. Umkhiqizo scalar enikezwa:

NN¹ = | N || N¹ | cos φ,

yingenxa

cosφ = NN¹ / | N || N¹ | = (AA¹ + VV¹ SS¹ +) / ((√ (A² + s² + V²)) * (√ (¹) ² + (H¹) ² + (S¹) ²)).

Kwanele cabanga ukuthi 0≤φ≤π.

Empeleni ezimbili izindiza aphambana, ifomu ezimbili engela (dihedral): φ ₁ futhi φ _2. sum yabo uyalingana π (φ ₁ + φ ₂ = π). Kepha cosines zabo, izindinganiso zabo ngokuphelele bayalingana, kodwa kunezimpawu ezingefani, okungukuthi, cos φ ₁ = -cos φ _2. Uma kule ndaba (0) esikhundleni A, B no C -A, -B futhi -C ngokulandelana, ezothando, sithola, ziyonquma indiza efanayo, i-engeli kuphela φ e kwesibalo cos φ = NN ¹ / | N || N ¹ | It sizobuyiselwange π-φ.

I kwesibalo indiza perpendicular

Ngokuthi perpendicular indiza, phakathi lapho engela 90 degrees. Ukusebenzisa obekuxoxwa ngenhla, singathola kwesibalo indiza perpendicular nezinye. Ake sithi sinalo izindiza ezimbili: Axe + ngu + CZ + D = 0, futhi + A¹h V¹u S¹z + D = 0. Singasho ukuthi zingabantu orthogonal uma cos = 0. Lokhu kusho ukuthi NN¹ = AA¹ + VV¹ SS¹ + = 0.

I kwesibalo indiza parallel

Lalibhekisela izindiza ezimbili okuhambisana nawo aqukethe akukho amaphuzu afana ngakho nawo.

Isimo izindiza parallel (zibalo zabo ziyefana esigabeni esandulele) ukuthi zithwala N futhi N¹, okuyizinto perpendicular kubo, collinear. Lokhu kusho ukuthi uma kuhlangabezana nemibandela elandelayo proportionality:

A / ¹ = B / C = H¹ / S¹.

Uma imigomo ezilinganiselwe kuthiwa wandisa - A / ¹ = B / C = H¹ / S¹ = DD¹,

lokhu kubonisa ukuthi indiza idatha ye okufanayo. Lokhu kusho ukuthi isibalo Axe + ngu + CZ + D = 0 ne- + A¹h V¹u S¹z + D¹ = 0 ukuchaza indiza eyodwa.

Ibanga kusuka iphuzu ukuze indiza

Ake sithi ube P indiza onikelwa ngu (0). Kuyadingeka ukuthola ibanga ukusuka iphuzu nge izixhumanisi (hₒ, uₒ, zₒ) = Qₒ. , Udinga ukuletha kwesibalo indiza II ukubukeka evamile ukwenza kube:

(Ρ, v) = p (r≥0).

Kulokhu, ρ (x, y, z) iyona Vector engaba iphuzu lethu Q, esisogwini n p - n ubude perpendicular, eyakhishwa kusukela zero iphuzu, v - kuyinto iyunithi Vector, okuyinto ahlelwe isinqophiso.

Umehluko ρ-ρº engaba vector of Q iphuzu = (x, y, z), okuqondene n futhi engaba Vector ngephuzu inikezwe Q ₀ = (hₒ, uₒ, zₒ) kuyinto Vector ezinjalo, inani eliphelele projection zazo ku v lilingana d ibanga, okuyilona elidingekayo ukuze uthole kusukela Q = ₀ (hₒ, uₒ, zₒ) P:

D = | (ρ-ρ ^0, v) |, kodwa

(Ρ-ρ ^0, v) = (ρ, v ) - (ρ ^0, v) = p (ρ ^0, v).

Ngakho-ke kuvela,

d = | (ρ ^0, v) p |.

Manje kusobala ukuthi ukubala d ibanga kusuka ₀ kuya Q indiza P, kubalulekile ukusebenzisa yokubuka ejwayelekile indiza equation, shift kwesokunxele p, futhi indawo yokugcina ye x, y, z esikhundleni (hₒ, uₒ, zₒ).

Ngakho, sithola inani eliphelele le nkulumo okuholela ukuthi liyadingeka d.

Ukusebenzisa kwemigomo ulimi, sithola esobala:

d = | Ahₒ Vuₒ + Czₒ | / √ (A² + V² + s²).

Uma esibekiwe iphuzu Q ₀ kuyinto ngakolunye uhlangothi P indiza njengoba umsuka ke phakathi Vector ρ-ρ ⁰ futhi v kuyinto engela obtuse, ngaleyo ndlela:

d = - (ρ-ρ ^0, v) = (ρ ^0, v) -p> 0.

Esimeni lapho iphuzu Q ₀ ngokuhlanganyela umsuka elise ohlangothini olufanayo U, i-engeli acute idaliwe, okungukuthi:

d = (ρ-ρ ^0, v) = p - (ρ ^0, v)> 0.

Umphumela uba ukuthi esimweni sangaphambili (ρ ^0, v)> p, okwesibili (ρ ^0, v)

Futhi indiza tangent yayo kwesibalo

Ngokuphathelene indiza kuya ebusweni endaweni tangency Mº - indiza equkethe zonke tangent kungenzeka ijika ungenisa ngalowo iphuzu phezu.

Nge leli fomu kobuso kwesibalo F (x, y, z) = 0 kule ndaba we Mº indiza tangent tangent iphuzu (hº, uº, zº) kungaba:

F _x (hº, uº, zº) (hº x) + F _x (hº, uº, zº) (uº y) + F _x (hº, uº, zº) (z-zº) = 0.

Uma ebusweni isethwe ngokucacile z = f (x, y), khona-ke indiza tangent sichazwe kwesibalo:

z-zº = f (hº, uº) (hº x) + f (hº, uº) (y uº).

Empambana izindiza ezimbili

Ngo isikhala ngakuthathu kuyinto Ukudidiyela uhlelo (unxande) Oxyz, unikezwa izindiza ezimbili P 'futhi P' ezidlulana futhi kungahambisani. Njengoba noma yimuphi indiza, lowo unxande Ukudidiyela uhlelo kuchazwe ezothando jikelele, sicabanga ukuthi n 'futhi n "ichazwe ngu zibalo A'x + V'u S'z + D' = 0 no" + B x '+ y nge "z + D" = 0. Kulokhu esinayo n evamile '(A', B ', C) we P indiza' kanye n evamile "(A", B ", C") we P indiza '. Njengoba indiza yethu kungukuthi zifana futhi kungahambisani ke lezi zithwala kungukuthi collinear. Ukusebenzisa ilimi le-IiMbalo, sinale isimo singatlolwa bunjesi: n '≠ n "↔ (A', B ', C') ≠ (λ * Futhi", λ * Ngo ", λ * C"), λεR. Vumelani emgceni locondzile elisemandleni empambana P 'futhi P ", uzobe okhonjiswe incwadi, kulokhu a = P' ∩ P".

futhi - umugqa ehlanganisa sebuningini amaphuzu (ezivamile) ezindizeni P 'futhi P ". Lokhu kusho ukuthi izixhumanisi iliphi iphuzu okuqondene umugqa a, kumelwe kanyekanye ukwanelisa ezothando A'x + V'u S'z + D '= 0 no "x + B' + C y" z + D "= 0. Lokhu kusho ukuthi izixhumanisi iphuzu kuyoba isixazululo ethile zibalo ezilandelayo:

Umphumela uba ukuthi ikhambi (jikelele) zalesi simiso zibalo ziyonquma izixhumanisi ngayinye amaphuzu emgqeni lapho uyokwenza njengoba iphuzu empambana P 'futhi P ", futhi ukunquma umugqa endaweni Ukudidiyela uhlelo Oxyz (unxande) isikhala.